Integration Sum Help

Evaluate

n=1,3,5,7....e(2n1)0n+1xeex(x+1)!dx \Large{\sum _{ n=1,3,5,7.... }^{ \infty }{ \frac { { e }^{ \left( 2n-1 \right) } }{ \int _{ 0 }^{ n+1 }{ \frac { { x }^{ e }{ e }^{ x } }{ \left( x+1 \right) ! } dx } } } }

Note by Department 8
3 years, 9 months ago

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@Jake Lai

Department 8 - 3 years, 9 months ago

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Very unlikely this series has a closed form. Factorials/gamma functions in integrals, especially in denominators of integrals, are difficult to evaluate in general.

It is best to proceed in the direction of a motivating problem which applies integration in its solution, or to build a problem on the back of techniques already well-studied.

Jake Lai - 3 years, 9 months ago

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@Jake Lai I made it accidently

Department 8 - 3 years, 9 months ago

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@Pi Han Goh

Department 8 - 3 years, 9 months ago

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What have you tried? What makes you think there's a closed form?

Pi Han Goh - 3 years, 9 months ago

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I don't know I just made this problem and asked my school teacher, she said it was pretty tough. So I asked for help.

Department 8 - 3 years, 9 months ago

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Not all series /integrals has a nice closed form.

Pi Han Goh - 3 years, 9 months ago

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@Pi Han Goh Sorry

Department 8 - 3 years, 9 months ago

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@Pi Han Goh try this

Department 8 - 3 years, 9 months ago

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@Department 8 See the report. Your problem is flawed.

Pi Han Goh - 3 years, 9 months ago

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I'm currently busy now. Ty using the infinite product of gamma(x+2)

Aditya Kumar - 3 years, 9 months ago

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@Jonas Katona

Department 8 - 3 years, 9 months ago

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